Find the slant asymptote and the vertical asymptote. (If an answer does not exist, enter DNE.) $$ r(x)=\frac{9 x-x^{2}}{8 x-8} $$ slant asymptote vertical asymptote

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Function by following steps:
- step0: Find the vertical asymptotes:
  $$r\left(x\right)=\frac{9x-x^{2}}{8x-8}$$
- step1: Evaluate the limit:
  $$\lim _{x\rightarrow 1}\left(\frac{9x-x^{2}}{8x-8}\right)$$
- step2: Evaluate the left-hand and the right-hand limits:
  $$\begin{align}&\lim _{x\rightarrow 1^{-}}\left(\frac{9x-x^{2}}{8x-8}\right)\&\lim _{x\rightarrow 1^{+}}\left(\frac{9x-x^{2}}{8x-8}\right)\end{align}$$
- step3: Evaluate the left-hand limit:
  $$\begin{align}&-\infty\&+\infty\end{align}$$
- step4: Limit does not exist:
  $$\textrm{The limit does not exist}$$
- step5: $$x=1$$ is a vertical asymptote$$:$$
  $$\begin{align}&x=1\textrm{ }\textrm{is a vertical asymptote}\end{align}$$
- step6: List all vertical asymptotes of the function:
  $$\begin{align}&x=1\end{align}$$
Find the oblique asymptotes of $$ r(x)=\frac{9 x-x^{2}}{8 x-8} $$.
Function by following steps:
- step0: Find the oblique asymptotes:
  $$r\left(x\right)=\frac{9x-x^{2}}{8x-8}$$
- step1: Evaluate the limits $$\lim _{x\rightarrow +\infty}\left(\frac{r\left(x\right)}{x}\right)$$ and $$\lim _{x\rightarrow -\infty}\left(\frac{r\left(x\right)}{x}\right):$$
  $$\begin{align}&\lim _{x\rightarrow +\infty}\left(\frac{\frac{9x-x^{2}}{8x-8}}{x}\right)\&\lim _{x\rightarrow -\infty}\left(\frac{\frac{9x-x^{2}}{8x-8}}{x}\right)\end{align}$$
- step2: Calculate:
  $$\begin{align}&-\frac{1}{8}\&-\frac{1}{8}\end{align}$$
- step3: Evaluate the limit $$\lim _{x\rightarrow +\infty}\left(r\left(x\right)-ax\right):$$
  $$\begin{align}&\lim _{x\rightarrow +\infty}\left(\frac{9x-x^{2}}{8x-8}+\frac{1}{8}x\right)\&\lim _{x\rightarrow -\infty}\left(\frac{9x-x^{2}}{8x-8}+\frac{1}{8}x\right)\end{align}$$
- step4: Calculate:
  $$\begin{align}&1\&1\end{align}$$
- step5: Write the oblique asymptote according to limits value:
  $$\begin{align}&r\left(x\right)=-\frac{1}{8}x+1\textrm{ }\textrm{is an oblique asymptote}\&r\left(x\right)=-\frac{1}{8}x+1\textrm{ }\textrm{is an oblique asymptote}\end{align}$$
- step6: The function has one oblique asymptote:
  $$r\left(x\right)=-\frac{1}{8}x+1$$
The results for the asymptotes of the function $$ r(x) = \frac{9x - x^2}{8x - 8} $$ are as follows:

1. **Slant Asymptote**: 
   The slant (or oblique) asymptote is given by the equation:
   $$
   r(x) = -\frac{1}{8}x + 1
   $$

2. **Vertical Asymptote**: 
   The vertical asymptote occurs at:
   $$
   x = 1
   $$

Thus, the slant asymptote is $$ -\frac{1}{8}x + 1 $$ and the vertical asymptote is $$ x = 1 $$.
Slant asymptote: $$ -\frac{1}{8}x + 1 $$ Vertical asymptote: $$ x = 1 $$

Find the slant asymptote and the vertical asymptote. (If an answer does not exist, enter DNE.) \[ r(x)=\frac{9 x-x^{2}}{8 x-8} \] slant asymptote vertical asymptote

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